AMAZON Coding Question – Solved

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As an operations engineer at Amazon, you are responsible for organizing the distribution of n different items in the warehouse. The size of each product is provided in an array `productSize`, where productSize[i] represents the size of the i-th product. You construct a new array called `variation`, where each element `variation[i]` is the difference between the largest and smallest product sizes among the first i products. Mathematically, this is defined as: variation[i] = max(productSize[0], productSize[1], ..., productSize[i]) - min(productSize[0], productSize[1], ..., productSize[i]) Your goal is to arrange the products in a way that minimizes the total variation, i.e., the sum of variation[0] + variation[1] + ... + variation[n-1]. Determine the minimum possible value of this sum after you have reordered the products. Example: n = 3 productSize = [3, 1, 2] By reordering the products as productSize = [2, 3, 1]: - variation[0] = max(2) - min(2) = 2 - 2 = 0. - variation[1] = max(2, 3) - min(2, 3) = 3 - 2 = 1. - variation[2] = max(2, 3, 1) - min(2, 3, 1) = 3 - 1 = 2. The sum is variation[0] + variation[1] + variation[2] = 0 + 1 + 2 = 3. This is the minimum possible total variation after rearranging. Function Description: Complete the function `minimizeVariation` in the editor below. `minimizeVariation` has the following parameter(s): - int productSize[n]: The size of each product. Returns: - int: The minimum possible total variation after rearranging the array productSize. Constraints: - 1 ≤ n ≤ 2000 - 1 ≤ productSize[i] ≤ 10^9 Sample Input 0: productSize = [4, 5, 4, 6, 2, 6, 1, 1] Sample Output 0: 16 Explanation: By reordering the products as productSize = [1, 1, 2, 4, 4, 5, 6]: - variation[0] = max(1) - min(1) = 1 - 1 = 0. - variation[1] = max(1, 1) - min(1, 1) = 1 - 1 = 0. - variation[2] = max(1, 1, 2) - min(1, 1, 2) = 2 - 1 = 1. - variation[3] = max(1, 1, 2, 4) - min(1, 1, 2, 4) = 4 - 1 = 3. - variation[4] = max(1, 1, 2, 4, 4) - min(1, 1, 2, 4, 4) = 4 - 1 = 3. - variation[5] = max(1, 1, 2, 4, 4, 5) - min(1, 1, 2, 4, 4, 5) = 5 - 1 = 4. - variation[6] = max(1, 1, 2, 4, 4, 5, 6) - min(1, 1, 2, 4, 4, 5, 6) = 6 - 1 = 5. The minimum total variation is variation[0] + variation[1] + variation[2] + variation[3] + variation[4] + variation[5] + variation[6] = 0 + 0 + 1 + 3 + 3 + 4 + 5 = 16.

Asked in: AMAZON

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Solution


from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10**6)
def minimizeVariation(productSize):
// ... rest of solution available after purchase

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Explanation


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To solve this problem effectively, you need to begin by deeply understanding the behavior of the `variation` array and how it is computed. For each index `i`, `variation[i]` is defined as the difference between the maximum and minimum values from index 0 to index i in the array. This means the variation array captures the running spread of the product sizes as you process the array from left to right.

The objective is to rearrange the original product sizes such that the cumulative sum of all these variations is minimized. One way to interpret this is to think about how quickly or slowly the maximum and minimum values evolve as you move through the array. Since every new element has the potential to become a new max or new min for the subarray from 0 to i, the order in which elements appear affects the range dramatically.

Now, let’s think about what kind of arrangement would cause minimal expansion of the max-min range over time. If you start with extreme values early on, then subsequent elements can cause the range to increase rapidly. For instance, if the first element is very small and later you insert a very large value, or vice versa, the difference between the running max and min becomes large quickly. That’s undesirable since we want to minimize the cumulative variation.

Instead, you want to build up the max and min slowly so that the variation doesn’t grow too fast. A useful way to approach this is to think of the values in sorted order. If the smallest values appear earlier and the larger values gradually appear later, the running max and min evolve more gradually. This results in smaller differences between them for each prefix of the array.

So one strong intuition is that sorting the array in ascending order can help keep the running max and min differences under control. Let’s try to understand why that helps.

When you sort the array and process elements from the smallest to largest, the running minimum is fixed from the beginning because the smallest number is at the front. The running maximum, however, grows slowly, because each next number is slightly larger than the previous. This leads to a slow increase in variation[i] values. At every step, the max - min is increasing, but only by small increments.

Contrast this with a random or descending order where the running max could be large early and the running min could come much later. This would mean that the range is already large early on, and remains large for most of the variation array, contributing significantly to the total sum.

Another important detail is that the variation at index 0 is always 0 regardless of the arrangement, because the prefix of length 1 will always have max and min equal to that one value. So, the contribution to the variation sum really starts from index 1 onward.

Now consider what happens in the case of duplicates. If the array contains many repeated elements, then grouping similar values together helps. For instance, placing all the smallest repeated values first ensures the minimum remains constant for a longer stretch, while the maximum only starts growing later. This again contributes to keeping the variation sum small.

To validate this strategy, look at the provided example: when the productSize is reordered in ascending order, the sum of variations is minimized. This suggests that a sorted order is likely to be optimal or near-optimal for this problem. The core reasoning behind this is to minimize how rapidly the running max diverges from the running min.

So, the high-level strategy becomes:
1. Understand that variation[i] depends on the max and min of the first i+1 elements.
2. Observe that the sooner the difference between max and min grows, the higher the cumulative variation will be.
3. Conclude that to slow down the growth of this difference, it is advantageous to process the values from the smallest to the largest.
4. Recognize that sorting the product sizes in ascending order naturally delays the increase in max values, while the minimum is fixed early on.

By executing this plan, you ensure that each new element only slightly increases the variation, leading to a minimized total. The key insight is how the ordering affects the running spread between max and min values, and how gradual growth in this spread keeps the cumulative variation low.

In summary, sorting the product sizes in increasing order leverages the properties of prefix max and min calculations to suppress rapid increases in variation, achieving the lowest total variation possible.
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